Introduction to Linear Polynomials | EOT

Question 6

The digits of a two-digit number differ by 3. If the digits are interchanged, and the resulting number is added to the original number, we get 143. Find both the numbers.

Solution

Understand the Question

Any two-digit number with tens digit $x$ and units digit $y$ is represented in expanded form as $10x + y$ .
When the digits are interchanged (reversed), the new number becomes $10y + x$ .
The phrase "digits differ by 3" means the difference between the two digits is $3$ $3$ , which leads to two possible cases:
- Case 1: $x - y = 3$ (tens digit is greater)
- Case 2: $y - x = 3$ (units digit is greater)
We form linear equations from the given conditions and solve for both possible numbers.

Step 1 · Form Equation from Sum of Numbers

Let the tens digit be $x$ and the units digit be $y$ .

$\text{Original number} = 10x + y$ $\text{Interchanged number} = 10y + x$

Given that the sum of the original number and the reversed number is $143$

\begin{aligned} (10x + y) + (10y + x) &= 143 \\ 11x + 11y &= 143 \\ 11(x + y) &= 143 \\ x + y &= 13 \quad \dots (1) \end{aligned}

Step 2 · Find Numbers When Tens Digit is Greater

If the tens digit is greater than the units digit $x - y = 3 \quad \dots (2)$

Adding equations $(1)$ and $(2)$

\begin{aligned} (x + y) + (x - y) &= 13 + 3 \\ 2x &= 16 \\ x &= 8 \end{aligned}

Substitute $x = 8$ into equation $(1)$

\begin{aligned} 8 + y &= 13 \\ y &= 5 \end{aligned}

$\text{Original number} = 10(8) + 5 = 85$ $\text{Interchanged number} = 58$

Step 3 · Find Numbers When Units Digit is Greater

If the units digit is greater than the tens digit $y - x = 3 \quad \dots (3)$

Adding equations $(1)$ and $(3)$

\begin{aligned} (x + y) + (y - x) &= 13 + 3 \\ 2y &= 16 \\ y &= 8 \end{aligned}

Substitute $y = 8$ into equation $(1)$

\begin{aligned} x + 8 &= 13 \\ x &= 5 \end{aligned}

$\text{Original number} = 10(5) + 8 = 58$ $\text{Interchanged number} = 85$

Answer

The numbers are $85$ and $58$ .

Common Mistakes

Place Value Error: Writing the two-digit number as $xy$ (multiplication) instead of its expanded place-value form $10x + y$ .
Missing the Second Case: Assuming only $x - y = 3$ and neglecting $y - x = 3$ , which leaves out the valid pair $58$ .

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Q10

*10. The graph of a linear polynomial $p(x)$ passes through the points $(1, 5)$ and $(3, 11)$ .

(i) Find the polynomial $p(x)$ .

(ii) Find the coordinates where the graph of $p(x)$ cuts the axes.

(iii) Draw the graph of $p(x)$ and verify your answers.

Q11

*11. Let $p(x) = ax + b$ and $q(x) = cx + d$ be two linear polynomials such that:

(i) $p(0) = 5$ .

(ii) The polynomial $p(x) - q(x)$ cuts the $x$ -axis at $(3, 0)$ .

(iii) The sum $p(x) + q(x)$ is equal to $6x + 4$ for all real $x$ .

Find the polynomials $p(x)$ and $q(x)$ .

Q12

*12. Look at the first three stages of a growing pattern of hexagons made using matchsticks. A new hexagon gets added at every stage which shares a side with the last hexagon of the previous stage.

(i) Draw the next two stages of the pattern. How many matchsticks will be required at these stages?

(ii) Complete the following table.

(iii) Find a rule to determine the number of matchsticks required for the $n^{\text{th}}$ stage.

(iv) How many matchsticks will be required for the $15^{\text{th}}$ stage of the pattern?

(v) Can 200 matchsticks form a stage in this pattern? Justify your answer.

Q13

*13. Let $p(x) = ax + b$ and $q(x) = cx + d$ be two linear polynomials such that:

(i) The graph of $p(x)$ passes through the points $(2, 3)$ and $(6, 11)$ .

(ii) The graph of $q(x)$ passes through the point $(4, -1)$ .

(iii) The graph of $q(x)$ is parallel to the graph of $p(x)$ .

Find the polynomials $p(x)$ and $q(x)$ . Also, find the coordinates of the point where these lines meet the $x$ -axis.

Q14

*14. What do all linear functions of the form $f(x) = ax + a, a > 0$ , have in common?